# How to find the rank of matrix $A$?

This question is from my assignment in linear algebra and I am not able to make any significant progress on this.

Question: Let $$A$$ be a $$5\times 4$$ matrix with real entries such that $$Ax=0$$ iff $$x=0$$ where $$x$$ is a $$4\times 1$$ vector and $$0$$ is a null vector. Then the rank of $$A$$ is ?

I am sorry but I don't have any intuition regarding what should be done in this question.

I have been following the textbook Linear Algebra by Hoffman and Kunze.

Kindly help.

Forget matrices. Matrices are only the computional aspect of linear algebra, which in small dimensions up to $$3$$ is elementary geometry.

For example, see $$A$$ as the matrix of a linear map $$u:\mathbb R^4\to \mathbb R^5$$ in canonical basis.

Then you have $$\dim \ker u+\dim Im (u)=4$$ because of $$\begin{matrix}\mathbb R^4 & \overset{u}{\longrightarrow} & \mathbb R^5 \\\downarrow & & \uparrow \\ \mathbb R^4/\ker u & \overset{\approx }{\longrightarrow} & Im(u) \\\end{matrix}$$

Your hypothesis translates into the fact that $$\ker u=\{0\}$$

So, $$rank A:=rank (u):=\dim Im(u)=4.$$

• Why not write $A$ everywhere you write $u$? Apr 10 at 8:15
• To avoid a notation clash. $A=Mat(u; e,f)$, where $e,f$ are basis. It depends on $e$ and $f$ chosen. But rank is intrinsically linked to $u$. Apr 10 at 8:17
• In other words, $u$ is not $A$. A morphism between vector spaces $V$ and $W$ is not the matrix that represents it in basis of $V$ and $W$. It's just a useful calculation tool for $u$. Apr 10 at 8:24